change the cartesian integral into an equivalent polar integral:

- anonymous

change the cartesian integral into an equivalent polar integral:

- chestercat

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- anonymous

\[\int\limits_{\sqrt{2}}^{2}\int\limits_{\sqrt{4-y^{2}}}^{y} dx dy\]

- amistre64

|dw:1353881320734:dw|i think this is what it looks like at least

- amistre64

remember that x=rcos(t)
y=rsin(t)
and r=sqrt(x^2+y^2), but thats really already given as 2

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## More answers

- anonymous

I think I understand how to get pi/4 < theta < pi/2
But finding the bounds for r is throwing me off.

- amistre64

well, since r only really changes from 0 to 2; the bounds for r are 0 to 2

- amistre64

it r changed according to some function of t
the r would be bound from 0 to f(t)

- anonymous

according to the answers, r is bounded by 2 < r < 2csc(theta)

- amistre64

|dw:1353881730467:dw|

- anonymous

but at pi/2 r can be 2 and rad(2)

- amistre64

it might help to see the limits in this fashion to be able to see how to change them out
\[\Large \int_{y=\sqrt{2}}^{y=2}~~~\int_{x=\sqrt{4-y^{2}}}^{x=y} ~~~dx~dy\]

- anonymous

my book says to plug in x=rcos(theta) etc and magically gets the polar bounds. They dont show any intermediate steps so im really confused.

- amistre64

\[x= \sqrt{r^2-y^2}\]\[2cos(t) = \sqrt{2^2-(2sin(t))^2}\]\[4cos^2(t) = 4-4sin^2(t)\]\[cos^2(t) +sin^2(t)=1~\text{for any t}\]
\[x= y\]\[2cos(t) = 2sin(t)\]\[t=\frac{pi}4\]
something along those lines i believe

- amistre64

y = 2
2sin(t) = 2
sin(t) = 1; when t=pi/2
y=sqrt(2)
2sin(t)=sqrt(2)
sin(t)=sqrt(2)/2; when t=pi/4

- amistre64

i might need to see the steps in your books to refresh me memory

- anonymous

this is all that they show.

##### 1 Attachment

- anonymous

I just dont understand where they cot the csc from.

- amistre64

your d parts in that png are dydx ... not dxdy

- anonymous

well, thats a problem...
the question has it as dx dy

- anonymous

I dont think dy dx even works. Wouldnt it be y=y y=rad(4-y^2)

- amistre64

yeah, so the png might be a typo

- anonymous

ok so ignoring the png, would the r just be 0 to 2?

- amistre64

from what i can tell yes;
|dw:1353883154486:dw|

- amistre64

but then i might have been thinking of this backwards

- amistre64

so, r goes from 2 up to the line; its not the area inside the circle

- amistre64

if i were to attempt this; I would determine the area of the triangle and subtract out the area inside the circle part

- anonymous

But doesnt the outer circle ask that the Y bounds are 2 and rad(2) not the X?

- anonymous

oh you flipped it

- amistre64

i positioned the graph in a convenient manner; a swaped the x and y axis ,,, yes

- amistre64

so, what we want is a function in ploar from for the line y=2

- anonymous

well, even though the answer key messed up the dxdy, but 2-(pi/2) seems to be the right answer.

- amistre64

it is; and youd be able to see that y=2 converts to 2csc(t) if i could remember how to do it :)

- amistre64

got it
r^2 = x^2 + y^2 ; but y=2 is constant
r^2 = x^2 + 4 ; x = rcos(t)
r^2 = r^2cos^2(t) + 4
r^2-r^2cos^2(t) = 4
r^2(1-cos^2(t)) = 4
r^2 sin^2(t) = 4
r^2 = 4csc^2(t)
r = 2csc(t)

- amistre64

area of triangle = 2*2/2 = 2
area of sector = 4pi/8 = pi/2
area of shaded region = triangle - sector: 2 - pi/2

- anonymous

ok that makes sense..
one more question,
is it a coincidence that
y = 2
r sin(t) = 2
r = 2 / sin(t)
using 1/sin(t) = csc(t) trig identity
r = 2 csc(t)
?

- amistre64

not sure if its coincidence or not; but that seems to work out as well :)

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